Consider the dot product in a metric space between two vectors. Below I show that the dot product doesn't change by a co-ordinate transformation.

$g_{\mu \nu} A^{\mu} B^{\nu} = g_{\mu \nu} \frac{\partial x_{\mu}}{\partial x^{\prime}_{\alpha}} {A^\prime}^{\alpha} \frac{\partial x_{\nu}}{\partial x^{\prime}_{\beta}} {B^{\prime}}^{\beta}$

Since $g_{\mu \nu} \frac{\partial x_{\mu}}{\partial x^{\prime}_{\alpha}} \frac{\partial x_{\nu}}{\partial x^{\prime}_{\beta}} = g^{\prime}_{\mu \nu}$


$g_{\mu \nu} A^{\mu} B^{\nu} = g^{\prime}_{\mu \nu} {A^\prime}^{\alpha} {B^{\prime}}^{\beta}$

And if the dot product is invariant so should be the angle between the two vectors.

So what's special about conformal transformations then? Kindly tell me if I've done anything wrong here.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Sep 18 '17 at 14:58

Yes, all coordinate transformations preserve angles between vectors because coordinate transformations are simply changes in the way a particular manifold gets mapped into $\mathbb{R}^n$. So a coordinate system for (an open subset of) a manifold $\mathcal{M}$ is a bijection $f: U\subseteq\mathcal{M}\to V\subseteq\mathbb{R}^n$. Given two coordinate systems defined by $f$ and $g$ which cover the same patches of $\mathcal{M}$ and $\mathbb{R}^n$ then a coordinate transformation between them is just a map $g \circ f^{-1}: V\to V$.

Obviously such a thing can't change anything about the manifold or its metric: it's just changing how we represent it. There are no coordinates in nature so changing coordinates can't change any physics, ever.

But a conformal transformation is not a coordinate transformation: it is a change in the metric itself. In general such a change can alter angles between vectors, because we can choose whatever metric we like for $\mathcal{M}$ so long as it meets the requirements to be a metric.

A restricted sort of such transformations however, do preserve angles: ones where $\mathbf{g}'(p) = \alpha(p)\mathbf{g}(p)$, where $\mathbf{g}$, $\mathbf{g}'$ are the old and new metrics, $p\in\mathcal{M}$ and $\alpha(p)$ is a function of $p$.

You can, more-or-less equivalently, talk about conformal transformations as being diffeomorphisms between different manifolds such that the metrics are related by a scalar factor (or, really, the pulled-back metric is related by a scalar factor to the 'native' metric). You can see that this is mostly the same thing but it allows you to think about conformal relationships between manifolds which are not obviously the same: Penrose diagrams are a good example of that.

Note I am talking here about conformal transformations (also called conformal mappings) as they're used in GR. I think the other answer is really talking about conformal transformations between coordinate systems which is a different (although related) thing. So I'm assuming here what the question is asking about and my assumption might be wrong.

  • $\begingroup$ There might be a problem with nomenclature (there is conflicting usage in texts). I think that in the context that OP is talking about, the transformation $g \to g' = \Omega^2 g$ is known as a "Weyl Transformation" whereas "Conformal Transformations" is a coordinate transformation as described by OP. $\endgroup$
    – Prahar
    Sep 17 '17 at 18:22
  • $\begingroup$ @Prahar: yes, and I'm now confused about what OP was really asking for. I've just answered based on the usage I've seen, and that might be wrong. $\endgroup$
    – user107153
    Sep 17 '17 at 19:00
  • $\begingroup$ @Prahar If a conformal transformation is as described by OP, does that mean the statement "conformal transformations preserve angles" is content-free? $\endgroup$
    – knzhou
    Dec 22 '18 at 16:17

Let $\mathcal{M}^d$ be a $d$-dimensional $(p,q)$ Riemannian manifold and let $f\colon \mathcal{M}^d\to\mathcal{M}^d$ be a coordinate transformation. Such transformation is said to be conformal if, by definition, $$g'_{\mu\nu}(x') = e^{\omega(x)}g_{\mu\nu}(x)$$ $x' = f(x)$ being the way the coordinates are affected by such transformation (to be intended component-wise, obviously). Working out the above definition one is led to the following set of possible transformations: $$x'^{\mu} = x^{\mu} + a^{\mu}\qquad \textrm{translations}$$ $$x'^{\mu} = \lambda x^{\mu}, \quad \lambda\in\mathbb{R}\qquad \textrm{dilations}$$ $$x'^{\mu} = \Lambda^{\mu}_{\nu}x^{\nu}\quad \Lambda\in\textrm{SO}(p,q)\qquad \textrm{Lorentz}$$ $$x'^{\mu} = \frac{x^{\mu}-b^{\mu}x^2}{1-2b\cdot x +b^2 x^2} \qquad\textrm{special conformal}$$ the above are the only collection of transformation satisfying the original requirement for the metric to only change by a (positive) scaling factor. In standard terminology one says that conformal transformations preserve the angles as it is obviously so given the multiplicative pre-factor that cancels out when calculating the scalar products and then dividing by the norm of the original vectors.

Below I show that the dot product doesn't change by a co-ordinate transformation.

Well, the scalar product is, by definition of scalar, left unchanged by a change of coordinates, there is no need of calculating anything.

  • $\begingroup$ "Well, the scalar product is, by definition of scalar, left unchanged by a change of coordinates, there is no need of calculating anything...." and "in a conformal transformation the scalar product changes by a multiplicative factor". How are these two statements consistent with your first paragraph where you state that a conformal transformation is nothing more than a change of coordinates (albeit a special one)? $\endgroup$
    – Prahar
    Sep 15 '17 at 20:53
  • $\begingroup$ @Prahar Ooops, sorry, I meant to say the "metric $g$", of course: I corrected it. $\endgroup$
    – gented
    Sep 15 '17 at 20:55
  • 1
    $\begingroup$ In the first sentence you seem to be defining a coordinate transformation as a bijection from a manifold to itself, which AFAIK is nonstandard. What I would take to be the standard definition would be the approach given in Wald, in which there is an atlas composed of charts. Your definition seems odd, since you're calling it a coordinate transformation, but it's actually a coordinate-free definition. In that sentence you also make it sound as though you're talking about any manifold, but the remainder of your question seems only applicable to Minkowski space. $\endgroup$
    – user4552
    Sep 15 '17 at 23:44
  • $\begingroup$ in a conformal transformation the metric changes by a multiplicative factor but so do the norms Hmm...so you seem to want to define a conformal transformation as a type of change of coordinates. But norms are scalars, so they're invariant under a change of coordinates, making this a contradiction. $\endgroup$
    – user4552
    Sep 15 '17 at 23:47
  • $\begingroup$ Under what I take to be the standard definitions, a conformal transformation isn't a change of coordinates, it's a pointwise rescaling of the metric. By that definition, we don't actually have general transformation rules for tensors, only for things like curvature tensors, which can be expressed in terms of the metric. For example, it doesn't make sense to talk about how the stress-energy tensor transforms under a conformal transformation. So under the definitions that I think of as standard, it doesn't make sense to say that a conformal transformation scales the norms of vectors by $\Omega$. $\endgroup$
    – user4552
    Sep 15 '17 at 23:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.